If 0 ln(absylum)/ln b. But how does this explain the limit tends to 0

Question

If 0<b<1 then the lim(b^n) = 0. 
Can anyone please explain why? I understand how n> ln(absylum)/ln b.  But how does this explain the limit tends to 0

zzr0ck3r · Answer

notice that if you multiply two fractions between 0 and 1 you get something very small
like (1/4)(1/4) = 1/16
now do that for ever (1/4)(1/4)(1/4).....(1/4)(1/4)(1/4)...... = 1/4^(a very big number) = 1/infinity = 0

zzr0ck3r · Answer

proving this with rational numbers is easy enough, but proving with irrational numbers would not be so easy, depending on your level....

anonymous · Answer

Consider proof by contradiction.  Consider that $a_{n-1}  < a_{n} $ for all $n$ and that $a_n > 0$ for all $n$.  This means the series must always be decreasing.  It can't diverge to positive infinity because it is never increasing, and it cant diverge to negative infinity because none of the terms are negative.  Suppose it converges to $0\lt c\lt 1$.  Well $\log_b(c) > 0$, since $b$ and $c$ are positive.  If we let $n=\lfloor \log_b(c) \rfloor$, then $n\leq \log_b(c)$ and $b^n \leq c$.  We know $b^{n+1} \lt b^n \leq c$ so the sequence must continue to decrease below $c$, thus it cannot converge to it.

anonymous · Answer

Thank you both so much!

anonymous · Answer

Oh, the last case to consider is that it doesn't converge, but it oscillates.  That can't happen because it is always decreasing.